Question

If $a,b,x,y$ be real numbers such that $a^2+b^2=81,x^2+y^2=121$ and $ax+by=99$ ,then find the set of all possible values of $ay-bx$

• A. $(0,\frac{9}{11}]$
• B. $(0,\frac{9}{11})$
• C. $\left\{0\right\}$
• D. $[\frac{9}{11},\infty )$

Answer 1

The correct option is C $\left\{0\right\}$

Given $a^2+b^2=81,x^2+y^2=121$ and $ax+by=99$

we know Cauchy Schwartz's inequality

$(a^2+b^2)(x^2+y^2)\le (ax+by{)}^2$

$\Rightarrow \left(81\right)\times \left(121\right)\le (99{)}^2$

$\Rightarrow 9801\le 9801$

but equality holds only if

$\frac{a}{x}=\frac{b}{y}\Rightarrow ay=bx$

$\Rightarrow ay-bx=0$

$\therefore$ the set of all values of $ay-bx$ is $\left\{0\right\}$